Prove $int_Rfg,dmleq|f|_p^{1-p/r}|g|_p^{1-q/r}(int_Rf^pg^q,dm)^{1/r}$, where $1leq pleqinfty$ and...
$begingroup$
Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$
integration measure-theory lebesgue-integral holder-inequality
edited Dec 27 '18 at 10:56
Blue
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
$endgroup$
add a comment |
$begingroup$
Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$
integration measure-theory lebesgue-integral holder-inequality
edited Dec 27 '18 at 10:56
Blue
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
$endgroup$
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11
add a comment |
$begingroup$
Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$
integration measure-theory lebesgue-integral holder-inequality
edited Dec 27 '18 at 10:56
Blue
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
$endgroup$
Let $f$, $g$ be positive real functions. And $f in L^p(R)$, $g in L^q(R)$, and $1 leqslant p,q <infty$. Then $fg in L^1(R)$ and
$$ int_R fg ,dm;leqslant; |f|_p^{1-p/r}|g|_p^{1-q/r}left(int_R f^pg^q ,dmright)^{1/r}$$
Where
$$1leqslant p leqslant +infty quadtext{and}quadfrac{1}{r}=frac{1}{p}+frac{1}{q}-1$$
integration measure-theory lebesgue-integral holder-inequality
integration measure-theory lebesgue-integral holder-inequality
edited Dec 27 '18 at 10:56
Blue
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
edited Dec 27 '18 at 10:56
Blue
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
edited Dec 27 '18 at 10:56
Blue
49.7k870158
edited Dec 27 '18 at 10:56
Blue
49.7k870158
edited Dec 27 '18 at 10:56
Blue
49.7k870158
49.7k870158
asked Dec 27 '18 at 10:24
nnMannnMan
464
asked Dec 27 '18 at 10:24
nnMannnMan
464
asked Dec 27 '18 at 10:24
nnMannnMan
464
464
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11
add a comment |
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11
add a comment |
1 Answer
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Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
add a comment |
$begingroup$
Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
add a comment |
$begingroup$
Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
Hints: I suppose $1+frac 1 r =frac 1 p +frac 1 q$. Write $fg$ as $f^{p/r}g^{q/r} f^{1-p/r}g^{1-q/r}$ and apply Holder with conjugate indices $r$ and $frac r {r-1}$. You will get $(int f^{p}g^{q})^{1/r}$ as one of the factors. Now applying another Holder to complete the proof is fairly straightfoward.
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
edited Dec 28 '18 at 8:20
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Dec 27 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
add a comment |
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
$begingroup$
Actually, the condition is $frac1r=frac1p+frac1q-1$.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:09
add a comment |
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$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Dec 27 '18 at 11:03
$begingroup$
Shouldn't the norm for $g$ be $q$? While you're correcting that, why not add some context? I have an answer, but unless there is more context, it would not be right to post it.
$endgroup$
– robjohn♦
Dec 28 '18 at 8:11